Problem: $\lim_{x\to 2}\dfrac{x-2}{1-\sqrt{3x-5}}=$
Solution: Substituting $x=2$ into $\dfrac{x-2}{1-\sqrt{3x-5}}$ results in the indeterminate form $\dfrac{0}{0}$. This doesn't necessarily mean the limit doesn't exist, but it does mean we have to work a little before we find it. Since we have a rational expression with a square root on our hands, let's try to re-write it using the method of rationalization. $\begin{aligned} &\phantom{=}\dfrac{x-2}{1-\sqrt{3x-5}} \\\\ &=\dfrac{x-2}{1-\sqrt{3x-5}}\cdot\dfrac{1+\sqrt{3x-5}}{1+\sqrt{3x-5}} \gray{\text{Rationalize the denominator}} \\\\ &=\dfrac{(x-2)(1+\sqrt{3x-5})}{1^2-(3x-5)} \\\\ &=\dfrac{\cancel{(x-2)}(1+\sqrt{3x-5})}{-3\cancel{(x-2)}} \gray{\text{Cancel out common factors}} \\\\ &=\dfrac{1+\sqrt{3x-5}}{-3} \text{, for }x\neq 2 \end{aligned}$ This means that the two expressions have the same value for all $x$ -values (in their domains) except for $2$. We can now use the following theorem: If $f(x)=g(x)$ for all $x$ -values in a given interval except for $x=c$, then $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)$. In our case, $\dfrac{x-2}{1-\sqrt{3x-5}}=\dfrac{1+\sqrt{3x-5}}{-3}$ for all $x$ -values in the interval $(1.5,2.5)$ except for $x=2$. Therefore, $\lim_{x\to 2}\dfrac{x-2}{1-\sqrt{3x-5}}=\lim_{x\to 2}\dfrac{1+\sqrt{3x-5}}{-3}=-\dfrac{2}{3}$. (The last limit was found using direct substitution.) [I want to see how this looks graphically!] In conclusion, $\lim_{x\to 2}\dfrac{x-2}{1-\sqrt{3x-5}}=-\dfrac{2}{3}$.